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EVEN06 - PART II CHAPTER 6 Applications of Integration...

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Unformatted text preview: PART II CHAPTER 6 Applications of Integration Section 6.1 Section 6.2 Section 6.3 Section 6.4 Section 6.5 Section 6.6 Section 6.7 Area of a Region Between Two Curves . . . . . . . . . . 264 Volume: The Disk Method . . . . . . . . . . . . . . . . . 271 Volume: The Shell Method . . . . . . . . . . . . . . . . 278 Arc Length and Surfaces of Revolution . . . . . . . . . . 282 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 Moments, Centers of Mass, and Centroids . . . . . . . . . 290 Fluid Pressure and Fluid Force . . . . . . . . . . . . . . . 297 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 Review Exercises Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 CHAPTER 6 Applications of Integration Section 6.1 Area of a Region Between Two Curves Solutions to Even-Numbered Exercises 2 2 2. A 2 2x 5 x2 2x 1 dx 2 x2 4 dx 1 1 4. A 0 x2 x3 dx 6. A 2 0 x 1 3 x 1 dx 1 3 8. 1 1 x2 y x2 1 dx 10. 2 y 7 x3 3 x x dx 3 4 12. 4 sec2 x y cos x dx 2 6 5 4 3 x 2 (3, 6) 3 ( (2, 2 ) 3 2 −π,2 4 2 ) π ( 4 , 2) −2 2 1 −1 (3, 1) x 4 5 6 7 π (− 4 , 22 ) (π , 22 ) 4 −π 4 −2 π 4 x 8 14. f x gx A 1 2 2 1 2x 16. A 2 10 8 2 1 x 2 7 x 2 3 xx 8 10 dx 8 8 dx x 32 x 8 7x2 4 112 Matches (a) y x3 8 64 10x 2 80 1 7 20 18 4 y 3 (2, 9) (0, 2) 8 6 (8, 6) 1 (4, 0) x 1 2 3 4 2 (2, 9 ) 2 (8, 0) x 2 4 6 10 264 S ection 6.1 18. The points of intersection are given by x2 4x x2 1 3x x2 3 Area of a Region Between Two Curves 265 20. The points of intersection are given by: x2 4x x3 2 x x 2 0, 3 x 0 1 0 when x 3x when x 1 x 0, 3 A 1 dx 3 fx 0 3 g x dx 4x 3x d x 2 x x3 3 2 dx 32 x 2 3 0 A 0 3 x2 x2 0 4x 3x dx 3x 2 23 0 x2 0 3 x 3 9 y 6 5 4 3 2 3 x2 0 y 6 5 4 3 9 2 27 2 9 2 (3, 5) (3, 4) (0, 2) (0, 1) x 1 2 3 4 5 6 −1 x 1 2 3 4 5 5 22. A 1 y 1.0 0.8 0.6 0.4 0.2 1 x2 (1, 1) 0 dx 1 x 5 1 4 5 (5, 0.04) x 1 2 3 4 5 24. The points of intersection are given by 3 x x 1 1 2x 2 1 x x 0 1 1 3 y x3 3x2 3x 1 1 (2, 1) (1, 0) x 2 x3 x x2 xx 1 3x 2 3x 2x x 1 x 1 0 0⇒x 3 0, 1, 2 (0, −1) A 2 0 x 1 1 dx 1 43 0 2 2 x2 2 1 2 3 x 4 0 3 4 1 2 266 Chapter 6 Applications of Integration 26. The points of intersection are given by: 2y yy 3 y2 3 fy y (−3, 3) y 3 0 when y g y dy 0, 3 A 0 3 1 (0, 0) 2y 0 3 y2 y2 dy y dy 32 y 2 13 y 3 3 0 −3 −2 −1 x 1 3y 0 9 2 3 y 28. A 0 3 0 fy y 16 3 g y dy 4 3 y2 y2 3 0 dy 12 ( 3 ,3 7 ) 2 1 2 16 0 2y d y 7 1.354 −1 1 x 1 2 3 16 y2 0 4 1 30. A 0 4 4x 4 4 2 4 ln 2 x y dx (1, 4) 1 x 0 3 (0, 2) 4 ln 2 1 −1 x 1 3 1.227 32. The point of intersection is given by: x3 2x x3 1 34. The points of intersection are given by: x4 1 A x2 x2 2 1 1 2x 0 when x g x dx 2x 2 4 2x 2 0 2 2x 2 0 when x x4 x4 dx x5 5 2 0 0, ± 2 A 1 1 fx x3 1 1 2 2 0 2x 2 d x 2x 1 dx 1 x4 4 2x d x 1 4x 2 4x 3 3 x 1 3 x 1 2 2 128 15 Numerical Approximation: 2.0 y Numerical Approximation: 8.533 10 (−2, 8) (2, 8) (−1, 2) −4 4 −2 (1, 0) −2 −1 −1 −2 x 2 (0, 0) (1, −2) S ection 6.1 36. f x x4 4x 2, g x x3 4x Area of a Region Between Two Curves y 267 The points of intersection are given by: x4 x4 xx 0 g4 3 4x 2 4x 2 x3 0 4x −4 −3 −1 x 1 3 4 x3 1x 4x 2 2x 0 when x 1 2, 0, 1, 2 2 −3 −4 f A 2 x3 248 30 37 60 4x 53 60 x4 293 30 4x 2 d x 0 x4 4x 2 x3 4x d x 1 x3 4x x4 4x2 d x Numerical Approximation: 8.267 3 0.617 0.883 9.767 4 38. A 0 6x x2 1 1 0 dx 3 3 40. A x 0 2 ( 3, ) 0 −1 −1 9 5 4 4 x dx x 3.434 3 ln x 2 3 ln 10 6.908 (0, 0) 5 −1 5 Numerical Approximation: 6.908 6 −1 1 42. A 2 cos 2x 1 sin 2x 2 3 4 y sin x dx 6 44. A 0 2 2 42 x 2 4x 4x 4 4 1 4 4 4 sec sec x x tan dx 4 4 x 4 1 cos x 2 0 3 2 0 33 4 1.299 2 2 y 4 3 2 1 24 2 2 2 2 4 ( −π 2 π, 1 62 ) x 2.1797 π 6 −1 (− π , −1) 2 (1, 2) x 1 46. From the graph we see that f and g intersect twice at x and x 1. 1 y 0 48. A 0 2 sin x 2 cos x cos 2x 1 sin 2x 2 (π, 1) 5 4 0 dx 4 0 A 0 1 gx 2x 0 f x dx 3 (1, 3) 2 1 3x dx 1 0 −1 (0, 1) 2 − x2 21 x 1x 3 ln 3 (0, 1) 4 x 1 2 −2 1 ln 3 0.180 268 Chapter 6 5 Applications of Integration 50. A 1 4 ln x x 5 2 1 0 dx 2 ln 5 2 52. (a) y 5.181 (b) A 0 x e x, y 1 0, x 0, x 1 2 ln x 2 x e x d x. No, it cannot be evaluated by hand. (5, 1.29) (c) 1.2556 6 4 y = xe x 0 (1, 0) −2 −1 −1 2 x 54. F x 0 12 t 2 0 2 dt 13 t 6 x 2t 0 x3 6 2x (b) F 4 y (a) F 0 y 43 6 24 56 3 (c) F 6 y 36 12 48 20 20 16 16 12 12 8 8 4 4 x 1 2 3 4 5 6 1 2 3 4 5 6 x 20 16 12 8 4 x 1 2 3 4 5 6 y y 56. F y 1 4ex 2 dx 1 y 30 25 20 15 10 5 −1 8ex 2 1 8e y 2 8e 12 (a) F 0 (b) F 0 y 30 25 20 15 10 5 x 8 8e 12 3.1478 (c) F 4 y 30 25 20 15 10 5 8e2 8e 12 54.2602 1 2 3 4 −1 x 1 2 3 4 −1 x 1 2 3 4 4 58. A 2 4 2 9 x 2 7 x 2 6 12 7 dx x 6 4 5 dx 4 5 x 2 16 x 5 dx 7 x 2 21x 21 d x 6 72 x 4 y 6 4 2 4 7x 2 72 x 4 7 4 7 14 9 y = 2 x − 12 (4, 6) 5 y = − 2 x + 16 (6, 1) x 2 −2 −4 6 8 10 (2, −3) y=x−5 S ection 6.1 1 x2 x2 1 2x 1 2 Area of a Region Between Two Curves y 269 60. f x fx (0, 1) 3 4 1 2 1 4 1 f ( x) = 2 x +1 1 At 1, , f 1 2 Tangent line: y 1 2 1 x 2 1 . 2 1 x 2 1 x2 1 1 dx (1, 21 ) y=− 1x+1 2 x 1 2 1 3 2 2 1 or y 1 at x 0. arctan x x2 4 1 The tangent line intersects f x 1 A 0 1 x2 1 1 x 2 x 0 3 4 0.0354 62. Answers will vary. See page 417. 64. x 3 ≥ x on 1, 0 1 y x 3 ≤ x on 0, 1 Both functions symmetric to origin 0 1 (1, 1) (0, 0) x3 1 1 x dx 0 x3 0. 2 x2 2 x d x. −1 x 1 −1 Thus, 1 x3 1 x dx x (− 1, − 1) A 2 0 x3 dx x4 4 1 0 1 2 66. Proposal 2 is better, since the cummulative deficit (the area under the curve) is less. 9 68. A 2 0 9 2 x dx 9 0 9 b b 2 9x 9 9 x b bx 9 x2 2 b9 9 b 9 x2 2 9 y 81 0 12 9 b dx x dx 9 0 b 81 2 81 2 81 2 81 2 9 2 2.636 (− (9 − b), b) 6 (9 −b, b) x 3 6 2 0 −6 −3 −3 −6 29 b b 9 2 270 Chapter 6 n Applications of Integration y 70. lim →0 i 4 1 x i2 2 x 4i and x n 4x 4 is the same as n x3 3 2 2 5 f ( x) = 4 − x 2 where xi 2 3 2 4 2 x2 dx 32 . 3 (− 2, 0) −3 −1 1 1 −1 (2, 0) x 3 5 5 72. 0 7.21 0.26t 0.02t 2 7.21 0.1t 0.01t 2 d t 0 0.01t 2 0.01t 3 3 29 billion 12 0.16t d t 0.16t 2 2 5 0 $ 2.417 billion 74. 5% : P1 31 %: P2 2 5 893,000 e 0.05 t 893,000 e 0.035 t Difference in profits over 5 years: 893,000e 0.05t 0 893,000e 0.035t d t 893,000 e 0.05t 0.05 e 0.035t 0.035 5 0 893,000 25.6805 893,000 0.2163 Note: Using a graphing utility you obtain $193,183. 34.0356 $ 193,156 20 28.5714 76. The curves intersect at the point where the slope of y2 equals that of y1, 1. y2 0.08x2 k ⇒ y2 0.16x 1⇒x 1 .16 6.25 6.25 (a) The value of k is given by y1 6.25 k y2 0.08 6.25 3.125. 2 (b) Area k 2 0 6.25 y2 y1 d x 3.125 x2 2 x dx 6.25 0 2 0 0.08x 2 0.08x 3 3 2 3.125x 2 6.510417 1 16 2 13.02083 78. (a) A (b) V 80. True 6.031 2A 2 2 5.908 2 11.816 m 3 1 8 2 5.908 (c) 5000V 5000 11.816 59,082 pounds Section 6.2 Volume: The Disk Method 271 Section 6.2 2 Volume: The Disk Method 2 2. V 0 3 4 x2 2 dx 0 3 x4 8x 2 16 dx x5 5 8x 3 3 2 16x 0 256 15 x2 4 x2 2 2 4. V 0 9 x2 2 dx 0 9 9x x2 dx x3 3 3 6. 2 18 8 x2 x 4 16 8 ±2 V 2 2 2 2 4 2 0 x2 4 2 2 2 dx 0 x4 16 2x3 3 2x2 12x 12 dx 2 0 2 2 2 2 x5 80 128 2 80 32 2 3 132.69 24 2 448 2 15 x2 ⇒ x 16 0 4 4 8. y V 16 4 16 2 y2 4 10. V 1 y2 y5 5 459 15 2y 4 153 5 4y 2 d y 1 y4 4 1 8y 3 16y 2 d y y2 4 0 dy 0 16 y2 dy 16y 3 3 16y y3 3 0, x 2, r y 8 128 3 2 y2 y dy 2 4y y2 4 8 12. y 2x 2, y (a) R y V (b) R x 16 0 2x 2, r x 2 0 4x 5 5 2 0 4 0 y 8 6 V 0 4x 4 d x y 8 6 128 5 4 2 x 2 4 −4 −2 4 2 x 2 4 −4 −2 (c) R x V 8, r x 2 8 64 4x 4 15 x 5 2 0 2x 2 32 x 2 dx 4 0 (d) R y 4 x4 dx 2 2 8 y 2, r y 2 y 2 4 y 2 2 0 dy y dy 2 y2 4 8 0 64 0 2 V 0 8 32x 2 0 8x 2 x4 dx 4 0 4 83 x 3 y 896 15 4y y 8 4 2 32 y 3 16 3 6 6 4 2 x 2 4 4 2 x 4 −4 −2 −4 −2 272 14. y Chapter 6 6 2x 6 0 Applications of Integration x 2, y 2x 6 x 6 intersect at x 2 3, 3 and 0, 6 . (b) R x 6 0 (a) R x V x 2, r x 2x 4x 3 x4 y 8 6 6 2 2x 3 2x x2 x2 3, r x 2 x 3 2 6 dx 3 x2 x dx V 3 0 x 3 0 x4 3 9x 2 3x 3 36x d x 3 0 x4 243 5 15 x 5 4x 3 x4 y 8 3x 2 x3 9x 2 18x d x 0 3 15 x 5 18x 2 3 108 5 4 2 x 2 4 2 x 2 −6 −4 −2 −6 −4 −2 16. R x 4 2 x3 ,r x 2 4 x3 2 2 0 18. R x V 4, r x 3 4 2 sec x 4 sec x 2 4 0 3 dx V 0 2 dx x6 4 x7 28 128 28 2 0 16 0 4x3 x4 16 8 sec x dx 0 sec 2 x d x 3 16x 32 y 8 ln sec x 8 ln 2 8 ln 2 y 5 tan x 3 3 3 3 tan x 0 144 7 8 ln 1 27.66 0 0 (2, 4) 3 2 1 −1 −1 x 1 2 3 4 3 2 1 π 9 π 3 x 2π 9 4π 9 5π 9 20. R y V 6, r y 4 6 y y3 3 4 0 2 6 dy 368 3 y y 5 4 3 2 1 y 6 0 2 36y x 1 −1 2 3 4 5 S ection 6.2 6 ,ry y 6 2 6 Volume: The Disk Method 273 22. R y V 36 6 6 0 24. R x V 2 x4 2 x 2, x2 rx 2 0 6 y 1 2 dy 2 y 1 dy y2 1 y 6 2 x4 0 2 dx 2 0 4x 2 4x 3 3 x4 dx x5 5 2 0 2 36 36 36 y 6 5 4 y 35 6 13 3 2 ln y 2 ln 6 1 2 ln 3 2 3 2 128 15 2 ln 2 6 ln 3 241.59 y 12 13 3 2 1 −3 −1 x 1 2 3 −2 3 −3 2 1 x 1 2 3 4 5 26. R x V 3 x 8 0 1 x 8 , rx 2 0 dx 2 28. R x V e x 2, r x 4 0 dx 3 1 1 1 x 1 ex 0 4 22 9 0 x dx 8 ex dx 0 4 8 0 9 y 4 ex 0 e4 y 8 1 168.38 3 6 2 4 1 2 x 2 4 6 8 −2 x 2 4 6 4 30. V 0 4 0 4 x2 4 1 x 2 5x 5x2 2 2 x 16 dx 2 8 dx 4 8 4 x x2 4 5x2 2 5x 2 4 16 dx 8 1 x 2 2 dx 4 3 2 y (4, 2) x3 12 88 3 4 16x 0 x3 12 16x 4 −2 1 x 2 −1 4 6 8 10 56 3 48 274 32. y x V Chapter 6 9 9 5 Applications of Integration 0, x 2, x 3 9 8 7 6 5 4 3 2 1 y x 2, y y 9 0 5 y y dy 2 22 d y (2, 5) 5 0 x 123456789 5y 25 y2 2 25 2 2 5 0 25 2 3 34. V 0 y cos x 2 dx 2.4674 36. V 1 ln x 2 d x 3.2332 2 1 x 1 2 5 38. V 0 2 arctan 0.2x 2 dx 15.4115 40. A 3 4 1 3 4 1 2 1 4 b y d 42. V a A x dx or V c A y dy Matches (b) x 1 4 1 2 3 4 1 44. (a) 4 2 2 4 x −4 z (b) 4 2 z (c) 4 4 z 8 y x 8 8 y 8 x 16 y a < c < b. 46. R x V r x, r x h h 0 0 y r2 2 x dx h2 h 0 r y= r x h (h, r) r2 3 x 3h 2 r h3 3h 2 2 x h 12 rh 3 S ection 6.2 4 Volume: The Disk Method 4 275 48. x V r2 r y2, R y r2 y2 2 2 r2 dy y2, r y 0 50. (a) V 0 x 2 dx 0 x dx x2 2 4 8 0 Let 0 < c < 4 and set c h r r h 2 2 y dy y3 3 r3 3 r 2h 3r 2h r h x dx 0 x2 2 c 0 c2 2 4. c2 c r 2h h3 3 h3 h3 3 (b) Set 0 8 8 22 ry r3 2r 3 3 3 2r 3 y Thus, when x 2 2 , the solid is divided into two parts of equal volume. c x dx c2 2 8 (one third of the volume). Then 3 16 ,c 3 d 8 , c2 3 4 3 x dx 0 43 . 3 16 3 (two thirds r To find the other value, set of the volume). Then x h d2 2 16 , d2 3 32 ,d 3 32 3 46 . 3 The x-values that divide the solid into three parts of equal volume are x 4 3 3 and x 4 6 3. 22.2, 0 ≤ x ≤ 11.5 11.5 < x ≤ 15 15 y 8 6 52. y V 0.1x 3 2.95, 11.5 2.2x 2 10.9x 0.1x 3 0 2.2x 2 10.9x 22.2 11.5 2 dx 11.5 2.95 2 d x 15 4 2 0.1x 4 4 2.2x 3 3 10.9x 2 2 22.2x 0 2.95 2x 11.5 x 1031.9016 cubic centimeters 54. (a) First find where y b x2 x 2 4 8 12 16 b intersects the parabola: 4 16 x2 4 4b b 4 x2 4 2 2 4 44 b 5 z 24 4 b V 0 4 b dx bx 2 2 dx 2 4 b b 4 x2 4 2 dx 2 x 2 4 y 4 0 4 0 x2 4 2x 2 2x 3 3 128 3 b x4 16 b2 b 2x 4b 2 8b 8bx 32b 16 d x 4 x5 80 64 5 bx 3 6 32 b 3 16x 0 64 4b 2 64 b 3 512 15 —CONTINUED— 276 Chapter 6 Applications of Integration 54. —CONTINUED— 64 b 3 512 15 64 3 64 3 8 8 3 (b) graph of V b 120 4b 2 (c) V b Vb 8 8b 0⇒b 22 3 >0⇒b 8 is a relative minimum. 3 0 0 4 Minimum Volume is 17.87 for b 10 2.67 56. (a) V 0 fx 2 dx a 2 Simpson’s Rule: b V 3 3 (b) f x 6 10 0 2 10, n 2 10 2 2.6 2 2.1 2 4 1.9 2 2.1 4 2.35 4 2.85 2 2 2.9 2 4 2.7 2 2 2.45 2 4 2.2 2 2.3 2 178.405 0.00249x 4 186.83 cm 3 0.0529x 3 0.3314x 2 0.4999x 2.112 0 0 10 10 (c) V 0 f x 2 dx 186.35 cm 3 58. V 1 10 2 3 2 30 m3 60. 3 y (b) A x 1 bh 2 34 1 24 2 x2 x2 dx x3 3 2 2 x2 34 x2 1 −3 −1 x 1 3 2 V 3 2 4 3 4x −3 32 3 3 Base of Cross Section (a) A x 2 24 2 x2 x2 2 4 − x2 2 4 − x2 b2 44 2 24 x2 dx x 3 32 2 V 2 4 − x2 2 4 − x2 4 4x 128 3 2 4 − x2 —CONTINUED— S ection 6.2 60. —CONTINUED— 12 r 2 2 2 Volume: The Disk Method 277 (c) A x (d) A x 4 x2 2 1 bh 2 1 24 2 x2 x 2 dx 4 x2 4x 4 x3 3 2 2 2 2 4 4x x2 x3 3 2 2 x2 32 3 V 2 4 2 x2 dx 16 3 2 V 2 4 4 − x2 2 4 − x2 2 4 − x2 62. The cross sections are squares. By symmetry, we can set up an integral for an eighth of the volume and multiply by 8. Ay r b2 r2 r2 y2 dy y2 2 y x V 8 0 8 r2y 16 3 r 3 13 y 3 r 0 R2 r2 64. V R2 R2 r2 r2 R2 2 0 x2 r2 x3 3 2 r2 dx R R2 R2 R2 R2 r2 x r2 3 2 r2 32 x2 dx R2 0 r2 r 2 2 4 3 R2 3 r2 32 R2 − r 2 R 66. (a) When a When a (b) y A 1 1 1: x 2: x x 1 1 2 y y 2 1 represents a square. 1 represents a circle. y 1 a=2 a=1 x a 1a 1 −1 1 2 x a 1a dx 4 0 1 xa 1a dx −1 To approximate the volume of the solid, form n slices, each of whose area is approximated by the integral above. Then sum the volumes of these n slices. 278 Chapter 6 Applications of Integration Section 6.3 2. p x hx V 2 0 1 Volume: The Shell Method 4. p x x 8 2 x 1 1 x x1 x 0 hx x dx x2 d x 2 x2 2 x3 3 1 0 x2 x4 0 2 4 x2 dx x3 dx x4 4 2 4 x2 V 2 2 0 2 3 4x 2x 2 2 8 0 6. p x hx V 2 x 12 x 2 6 0 8. p x hx V 2 x 4 2 10. p x x 2 x 4 2 hx x3 dx 14 x 4 2 2x x4 0 2 13 x dx 2 324 0 4x 0 V 8 2 2 0 2x d x 2x 2 d x 23 x 3 2 0 x 4 y 18 15 12 9 6 3 −1 −3 46 2 2x 2 y 0 4x 2x 2 y 2 3 4 3 16 3 2 1 x 1 2 2 1 x 1 2 3 x 1 2 3 4 5 6 −2 −1 −1 12. p x hx V 2 x sin x x x 0 3 2 y sin x dx x 2 cos x 0 1 x π 4 π 2 3π 4 π 2 0 sin x d x 4 −1 14. p y hy V 2 4 y 2 0 p y ≥ 0 on y y2 2 y dy y2 dy 0 2 2, 0 y 16. p y hy V 2 y 16 4 y2 4 y y 16 0 4 y2 dy y3 dy y4 4 64 4 0 3 2 1 x 2 0 2 2 2y y2 y3 3 2 0 16y 8y2 128 −1 −2 −3 −4 4 8 12 2 8 3 2 2 128 S ection 6.3 18. p x hx V 2 0 2 Volume: The Shell Method 279 2 4x 2 x x 2 8x 0 2 20. p x x 2 6 x 4 x 4x 2x 2 hx V 2 x 4x 8x 2 83 x 3 2x 2 d x 2x 3 d x 14 x 2 2 0 6 0 4 x x dx x3 2 2 2 y 4 3 2 0 6x 1 2 4x3 2 dx 192 5 4x2 16 3 y 4 3 2 2 25 x 5 4 2 0 2 1 1 x 1 −1 x 1 3 2 3 4 5 −1 −2 22. (a) Disk Rx V 1 (b) Shell 10 ,r x x2 5 y 0 10 Rx V 2 x, r x 5 0 10 dx x2 10 x2 5 2 8 dx 4 6 4 2 x 1 5 100 1 x x dx −1 x −2 1 2 3 4 5 20 1 1 dx x 5 35 100 100 3 (c) Disk Rx V 1 3 20 1 ln x 1 20 ln 5 1 125 1 496 15 10, r x 5 10 10 200 x 5 1 10 x2 10 x2 2 102 100 3x3 dx 1904 15 y 24. (a) Disk Rx rx a (b) Same as part a by symmetry y a2 3 0 a2 3 a a x2 3 32 (0, a) (0, a) (a, 0) V 2 0 x2 3 3 dx 3a 4 3x 2 3 9 43 53 ax 5 3a 2 3x 4 3 9 23 73 ax 7 13 a 3 (− a, 0) (a, 0) x x a2 a 2x a3 x2 dx 13 x 3 a 0 (0, − a) 2 2 93 a 5 93 a 7 32 a 3 105 280 26. (a) Chapter 6 z 2 Applications of Integration (b) 2 x 5 z (c) 2 z 10 y 2 x −2 5 y x 5 5 y a<c<b 4 28. 2 0 x x dx 2 30. (a) 1 3 4 1 2 1 4 y represents the volume of the solid generated by revolving the region bounded by y x 2, y 0, and x 4 about the y-axis by using the Shell Method. 2 2 16 0 2y 2 dy 0 4 2 2y 2 dy 1 4 1 2 3 4 x 1 represents this same volume by using the Disk Method. y 1 (b) V 4 3 2 1 x 1 −1 2 3 4 5 2 0 x1 x3 dx 2.3222 Disk Method 32. (a) 4 3 2 1 y 34. y tan x, y 1 0, x 0, x 4 Volume Matches (e) y x 1 2 3 4 2 3 (b) V 2 1 2x dx 1 e1 x 1 19.0162 π 4 π 2 x 36. Total volume of the hemisphere is 1 2 Find x0 such that x0 4 3 r3 2 3 3 3 18 . By the Shell Method, p x x, h x 9 x 2. 6 6 2 0 x0 x9 9 0 x2 dx 12 y x2 2x d x x0 2 1 x 2 9 3 9 x 02 32 x2 3 2 18 0 2 9 3 x 02 3 2 −3 −2 −1 −1 −2 −3 1 2 3 18 9 18 2 3 18 2 3 2.920 1.460. x0 Diameter: 2 9 S ection 6.3 r Volume: The Shell Method 281 38. V 4 r r R x r2 r2 x2 dx 2 x2 dx r 4R r 4 r x r2 r x 2 dx 4R 2 2 r 2R r2 2 22 r 3 x2 32 r b 40. (a) Area region 0 abn abn abn abn x 1 axn dx xn 1 a n1 a 1 bn n 1 n n n 1 1 1 1 1 b 0 (c) Disk Method: b V 2 0 x abn b axn dx xn 1 2a 0 xbn bn 2 x 2 bn 2 abn 2 dx b 0 1 abn 1 1 n n 1 2a 2a xn 2 n2 bn 2 n2 2 abn R1 n (b) lim R1 n n→ n→ 1 n abn 2 n n 2 n1 abn b n→ lim n n R2 n (d) lim R2 n n→ n→ n n2 b2 abn lim n n 2 n n 1 2 lim abn b n→ lim b2 abn , the graph approaches the line x 1. (e) As n → 4 42. (a) V 2 0 x f x dx 40 0 34 5800 50 4 10 45 2 20 40 4 30 20 0 2 20 3 121,475 cubic feet 40 20 40 x 1 x 2 12 x 2 25x 2 0 (b) Top line: y 50 x 0 0 40 40 x 20 50 d x 0 20 1 x⇒y 2 2x 40 1 x 2 20 ⇒ y 80 d x 80x dx 50 2x 80 Bottom line: y 20 V 2 0 20 2 20 x 40 2x 2x 2 2 0 50x dx 20 2 20 2 2 x3 6 26,000 3 2 32,000 3 2x 3 3 40 40x 2 20 2 121,475 cubic feet (Note that Simpson’s Rule is exact for this problem.) 282 Chapter 6 Applications of Integration Section 6.4 2. 1, 2 , 7, 10 (a) d (b) y y s 1 Arc Length and Surfaces of Revolution 4. y 1 2 3 2 2x3 2 3 0, 9 9x dx 9 7 4 x 3 4 3 7 10 2 2 10 y s 3x1 2, 9 1 0 2 1 27 1 4 3 2 9x 2 32 0 dx 5 x 3 7 10 1 2 823 27 1 54.929 6. y y s 32 x 2 x 3 4 , 1, 27 1 1 x1 3 3 2 8. y y x5 10 14 x 2 14 x 2 b 1 6x 3 1 2x 4 12 , 1, 2 2x 4 1 y 14 x 2 14 x 2 2 13 27 1 27 1 dx 1 y 2 x2 1 3 x2 27 dx s dx dx 1 2x 4 2 3 2 x2 3 1 2 11 3x 1 2 32 a 3 27 2 1 2 1 dx 3 2 103 2 22 x 3 3 1 23 28.794 1 dx 2x 4 1 6x 3 2 1 15 x 10 1x e 2 1x e 2 1x e 2 2 779 240 3.246 10. y y 1 y 2 e e x 12. (a) y , 0, 2 2 −3 x2 x 5 2, 2≤x≤1 x 2 e x , 0, 2 −5 s 0 1x e 2 2 2 e x x dx (b) 1 y y 2 2x 1 1 1 4x2 2 2 1 2 ex 0 e e x dx 12 e 2 1 e2 3.627 4x 4x 1 4x2 dx 1x e 2 2 0 L (c) L 5.653 14. (a) y 1 1 5 x ,0 ≤ x ≤ 1 (b) 1 y y 2 1 1 1 1 x 1 1 1 2 (c) L 1.132 y= 1 x+1 −1 −2 2 x 4 L 0 1 1 x 4 dx S ection 6.4 Arc Length and Surfaces of Revolution 283 16. (a) y cos x, 2 2 ≤x≤ y = cos x 2 (b) 1 y y 2 sin x 1 sin 2 x 2 (c) 3.820 −2 2 L 2 1 sin 2 x d x −2 18. (a) y ln x, 1 ≤ x ≤ 5 2 (b) 1 y y 2 1 x 1 5 (c) L 1 x2 1 1 dx x2 4.367 −1 6 L 1 −6 20. (a) x y 36 36 y2 , 0 ≤ y ≤ 3 x 2, 3 3 ≤ x ≤ 6 10 (b) dx dy 1 36 2 y 36 3 y2 12 2y (c) L 3.142 ! y2 1 y2 36 y2 dy y2 dy L −10 0 10 3 −5 0 6 36 Alternatively, you can convert to a function of x. y y L 3 3 36 dy dx 6 x2 x 36 1 x2 x2 36 6 dx x2 3 3 6 36 x2 dx 3.142. Although this integral is undefined at x 4 2 0, a graphing utility still gives L y 22. 0 1 s 1 d tan x dx dx 2 y = tan x 1 ( π , 1( 4 π 4 3π 8 x Matches (e) (0, 0) 24. f x (a) d (b) d x2 4 1 4 2, 0, 4 0 0 2 2 144 9 16 16 2 2 128.062 2 1 2 0 9 2 3 2 2 25 0 2 4 3 2 144 25 2 160.151 4 (c) s 0 1 4x x 2 4 2 dx 159.087 (d) 160.287 284 Chapter 6 Applications of Integration 1 and L1 x dx dy e 26. Let y ln x, 1 ≤ x ≤ e, y 1 1 1 d x. x2 1 1 Equivalently, x e y, 0 ≤ y ≤ 1, e y, and L 2 0 1 e 2y d y 0 1 e 2x d x. Numerically, both integrals yield L 10 e x 20 20 x 20 2.0035 28. y y 1 y 2 31 e x 20 1x e 2 1 20 e 10 1x e 4 2 20 e e x 20 x 10 1x e 2 2 2 20 e x 20 s 20 1x e 2 ex 20 20 x 20 dx 20 1 2 20 e dx 10 e x 20 e x 20 20 20 e 1 e 47 ft Thus, there are 100 47 30. y y s 299.2239 4700 square feet of roofing on the barn. 693.8597 68.7672 cosh 0.0100333x 0.6899619478 sinh 0.0100333x 299.2239 1 0.6899619478 sinh 0.0100333x 2 d x 100 or a graphing utility.) 1480 (Use Simpson’s Rule with n x2 x 25 25 25 4 32. y y 1 y 2 25 34. y y S 25 2x 1 x 2 4 9 x2 x2 25 5 25 x 5 4 5 s x 4 3 , 4, 9 9 2x x 4 1 1 dx 9 32 4 1 dx x s 3 4 3 x 2 dx 4 8 3 x dx 2 1 2 5 arcsin 5 arcsin 1 2 4 5 7.8540 8 103 3 3 5 7.8540 53 2 171.258 arcsin Section 6.4 x 2 1 2 5 , 0, 6 4 6 Arc Length and Surfaces of Revolution 285 36. y y 1 y 2 38. y y S 9 2x x2, 0, 3 3 2 0 3 x1 1 0 4x2 dx 12 S 2 0 x 2 5 6 5 dx 4 x2 0 4 6 4x2 4x2 2 8x dx 3 1 373 32 0 2 8 95 6 1 117.319 40. y y 1 y 2 ln x 1 x x2 x2 e 42. The precalculus formula is the distance formula between two points. The representative element is xi 1 , 1, e x2 x2 x2 1 2 yi 2 1 yi xi 2 xi. S 2 1 e x 1 dx 22.943 2 1 dx 44. The surface of revolution given by f1 will be larger. r x is larger for f1. 46. y r2 y 1 y 2 x2 x r2 r2 r2 x2 r 48. From Exercise 47 we have: a x2 S 2 0 a rx dx r2 x2 2x d x r2 x2 a r 0 S 2 r r r2 r dx r x2 2 rx r2 r2 r x2 dx 2r 2 2r 2r r2 2r x2 0 2 4 r2 r r2 r2 a2 a2 r 2 r h where h is the height of the zone y r h r−h x x2 + y2= r2 286 Chapter 6 Applications of Integration 50. (a) We approximate the volume by summing 6 disks of thickness 3 and circumference Ci equal to the average of the given circumferences: 6 6 V i 1 ri 2 3 i 1 2 Ci 2 2 3 3 4 70 2 2 6 Ci 2 i 1 3 4 50 2 65.5 65.5 70 2 66 2 66 2 58 2 58 2 51 2 51 2 48 2 3 57.75 2 4 67.75 2 68 2 62 2 54.5 2 49.5 2 3 21813.625 5207.62 cubic inches 4 (b) The lateral surface area of a frustum of a right circular cone is s R S1 32 50 2 65.5 2 65.5 9 50 2 12 r . For the first frustum, r s 50 2 50 65.5 2 2 12 65.5 2 . h Adding the six frustums together, S 50 2 70 2 58 2 224.30 1168.64 (c) r 0.00401 y 3 20 R 2 12 65.5 66 51 9 9 9 15.5 2 4 2 7 2 65.5 2 66 2 51 2 202.06 58 48 70 9 8 2 3 2 4.5 2 2 12 2 12 2 12 9 9 2 12 2 12 208.96 208.54 174.41 150.37 18 0.1416 y 2 1.232 y 7.943 (d) V 0 18 r2 dy 2 ry 0 5275.9 cubic inches 1 r y 2 dy S 1179.5 square inches −1 −1 19 52. Individual project, see Exercise 50, 51. x2 9 y2 4 x2 ,0 ≤ x ≤ 3 9 x2 9 12 54. (a) 1 2 1 4 (b) y 1 x2 9 x2 9 y 2 2 1 2 1 1 2x Ellipse: y1 y2 2 2x 9 2x x2 9 3 1 x2 9 1 39 −5 2 5 L 0 81 4x2 dx 9x2 x2 + y = 1 94 −4 (c) You can...
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